[LeetCode 90] Subsets II

题目链接：subsets-ii

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

/**
 * 
		Given a collection of integers that might contain duplicates, S, return all possible subsets.
		
		Note:
		Elements in a subset must be in non-descending order.
		The solution set must not contain duplicate subsets.
		For example,
		If S = [1,2,2], a solution is:
		
		[
		  [2],
		  [1],
		  [1,2,2],
		  [2,2],
		  [1,2],
		  []
		]
 *
 */

public class SubsetsII {

//	19 / 19 test cases passed.
//	Status: Accepted
//	Runtime: 259 ms
//	Submitted: 0 minutes ago
	
	//时间复杂度O(2^n) 空间复杂度 O(n)
	//偷懒了，直接用了set容器，去除重复子集
	public Set<List<Integer>> subsets = new HashSet<List<Integer>>();
	public List<List<Integer>> subsetsWithDup(int[] num) {
        Arrays.sort(num);
        subsets(num, 0, new ArrayList<Integer>());
        return new ArrayList<List<Integer>>(subsets);      
    }
    public void subsets(int[] S, int step, List<Integer> subset) {
        if(step == S.length) {
        	subsets.add(subset);
        	return;
        }
        //num[step] 不加入子集中
        subsets(S, step + 1, new ArrayList<Integer>(subset));
        
        //num[step] 加入子集中
        subset.add(S[step]);        
        subsets(S, step + 1, new ArrayList<Integer>(subset));        
    }
    
	public static void main(String[] args) {
		// TODO Auto-generated method stub

	}

}

[LeetCode 90] Subsets II

原文：http://blog.csdn.net/ever223/article/details/44687995